#include <bits/stdc++.h>
using namespace std;
#include "peru.h"
#define ll long long
const ll p = 1e9+7;
int solve(int n, int k, int* v){
vector<pair<ll, int>> dp(n); // min effort, last idx of prev segment
function<pair<ll, int>(int)> get = [&](int idx) {
if (idx < 0) return make_pair(0ll, -1);
return dp[idx];
};
vector<int> mon_s;
// int last_max_idx = 0;
for (int i = 0; i < n; i++) {
while (!mon_s.empty() && v[mon_s.back()] <= v[i]) mon_s.pop_back();
// last_max_idx = min(last_max_idx, (int)mon_s.size());
// while (last_max_idx != mon_s.size() && mon_s[last_max_idx] <= i-k) last_max_idx++;
mon_s.push_back(i);
int max_val = 0;
for (int j = max(i-k+1, 0); j <= i; j++) max_val = max(max_val, v[j]);
pair<ll, int> curmin = {1e18, -1};
curmin = min(curmin, {get(i-k).first + max_val, max(i-k, -1)});
if (mon_s.size() != 1 && mon_s[mon_s.size()-2] >= i-k) curmin = min(curmin, {get(mon_s[mon_s.size()-2]).first + v[i], mon_s[mon_s.size()-2]});
if (mon_s.size() != 1 && dp[mon_s[mon_s.size()-2]].second >= i-k) curmin = min(curmin, dp[mon_s.size()-2]);
// pair<ll, int> curmin = {1e18, 0};
// int runmax = v[i];
// for (int j = i-1; j >= i-k; j--) {
// curmin = min(curmin, ({(ll)get(j).first + (ll)runmax, j}));
// if (j < 0) break;
// runmax = max(runmax, v[j]);
// }
dp[i] = curmin;
}
ll factor = 1;
ll res = 0;
for (int i = n-1; i >= 0; i--) {
res += ((dp[i].first%p)*factor) % p;
res %= p;
factor *= 23;
factor %= p;
}
return res;
}
